a 2 uF capacitor is connected in series with a 1.2 M ohms resistor and a 5 v battery for a long time. What is the current in the resistor 1s after disconnecting the battery?
Current starts out at V/R, at t = 0, but decays so that
I(t) = (V/R)*e^(-t/RC)
In this case, RC = 2.4 seconds
V/R = 4.17*10^-6 amps
I(t=1s) = 4.17*10^-6*e^-0.4167
= 2.75 microamps